A 250-gram pendulum bob is initially constrained at a 53 degree angle by two strings. When the horizontal string is cut the pendulum begins to swing. As the pendulum bob passes through its equilibrium position, what will be its velocity?
To calculate the velocity of the pendulum bob as it passes through its equilibrium position, we can use the principle of conservation of energy.
The gravitational potential energy at the highest point of the swing is given by:
PE = mgh
where m is the mass of the pendulum bob (0.25 kg in this case), g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the height from the lowest point of the swing to the highest point.
Since the pendulum bob is initially at a 53-degree angle, the height h can be calculated as follows:
h = L(1 - cosθ)
where L is the length of the pendulum string and θ is the angle in radians.
Converting 53 degrees to radians:
θ = 53° * (π/180°) ≈ 0.925 radians
Now, we need to determine the length L of the pendulum string. Since the mass of the bob and angle are given, we can make use of the fact that the tension in the string is equal to the weight of the bob at the highest point of the swing:
T = mg
where T is the tension in the string. At the highest point, the tension can be resolved into horizontal and vertical components:
T_horizontal = T * cosθ = mg * cosθ
T_vertical = T * sinθ = mg * sinθ
Since the horizontal string is cut, the tension in the horizontal direction is zero:
T_horizontal = 0
Thus, we have:
mg * cosθ = 0
cosθ = 0
θ = π/2 radians
Now, we can use the equation for height calculation:
h = L(1 - cosθ)
h = L(1 - cos(π/2))
h = L(1 - 0)
h = L
Therefore, the height h is equal to the length of the pendulum string L.
Now, we can calculate the gravitational potential energy at the highest point:
PE = mgh = (0.25 kg)(9.8 m/s^2)(L)
As the pendulum bob passes through its equilibrium position, all of the gravitational potential energy is converted to kinetic energy. Therefore, the kinetic energy at the equilibrium position is:
KE = PE
Since kinetic energy is given by:
KE = 0.5mv^2
where v is the velocity, we can equate the expressions for kinetic energy and gravitational potential energy:
0.5mv^2 = (0.25 kg)(9.8 m/s^2)(L)
Simplifying the equation, we can solve for v:
v = √[(2(0.25 kg)(9.8 m/s^2)(L))/(0.5m)]
v = √[(4.9 m^2/s^2)(L)]
Thus, the velocity of the pendulum bob as it passes through its equilibrium position is given by v = √[(4.9 m^2/s^2)(L)].